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Prove that [(a^2+b^2)/(a+b)]^(a+b)> a^a ...

Prove that `[(a^2+b^2)/(a+b)]^(a+b)> a^a b^b >{(a+b)/2}^(a+b)dot`

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`([a+a+...a "times"]+[b+b+....b" times"])/(a+b)ge [a^ab^b]^((1)/(a+b))`
`ge (a+b)/(((1)/(a)+(1)/(a)+....a " times")+((1)/(b)+(1)/(b)+.....b" times"))`
or ` (a^2+b^2)/(a+b)ge[a^ab^b]^((1)/(a+b))ge (a+b)/(1+1)`
or ` ((a^2+b^2)/(a+b))^(a+b) ge a^a b^b ge ((a+b)/(2))^(a+b)`
or ` [(a^2+b^2)/(a+b)]^(a+b) gt a^a b^b gt {(a+b)/(2)}^(a+b)`
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